Difference between revisions of "1987 IMO Problems/Problem 2"
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Thus we have proven that <math>[AKNM]=[ABC]</math>. | Thus we have proven that <math>[AKNM]=[ABC]</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Clearly, <math>AKLM</math> is a kite, so its diagonals are perpendicular. Furthermore, we have triangles <math>ABN</math> and <math>ALC</math> similar because two corresponding angles are equal. | ||
+ | |||
+ | Hence, we have <math>[AKNM] = \frac{1}{2} AN \cdot KM = \frac{1}{2} \frac{AB \cdot AC}{AL} \cdot KM.</math> Notice that we used the fact that a quadrilateral's area is equal to half the product of its perpendicular diagonals (if they are, in fact, perpendicular). | ||
+ | |||
+ | But in (right) triangle <math>AKL</math>, we have <math><LAB = <A/2</math>. Furthermore, if <math>Q</math> is the intersection of diagonals <math>AL</math> and <math>KM</math> we have <math>Q</math> the midpoint of <math>KM</math> and <math>KQ</math> an altitude of <math>AKL</math>, so | ||
+ | <cmath>\frac{KM}{2} = \frac{AK \cdot KL}{AL} = \frac{AL \sin <A/2 \cdot AL \cos <A/2}{AL} = \frac{AL \sin <A}{2},</cmath> | ||
+ | so <math>\frac{KM}{AL} = \sin <A</math>. Hence <cmath>[AKNM] = \frac{1}{2} \frac{AB \cdot AC}{AL} \cdot KM = \frac{1}{2} AB \cdot AC \sin <A = [ABC],</cmath> as desired. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Proceed as in Solution 2. To prove that <math>\frac{KM}{AL} = \sin A</math>, consider that | ||
+ | |||
+ | <cmath>AL = \frac{AK}{\sin <KLA} = \frac{AK}{\sin <AKM} = \frac{KM}{\sin A}</cmath> | ||
+ | |||
+ | via usage of definition of sine, equal angles in a right triangle if its altitude is drawn, and the Law of Sines. | ||
==See also== | ==See also== |
Latest revision as of 15:39, 21 September 2014
Problem
In an acute-angled triangle the interior bisector of the angle
intersects
at
and intersects the circumcircle of
again at
. From point
perpendiculars are drawn to
and
, the feet of these perpendiculars being
and
respectively. Prove that the quadrilateral
and the triangle
have equal areas.
Solution
We are to prove that or equivilently,
. Thus, we are to prove that
. It is clear that since
, the segments
and
are equal. Thus, we have
since cyclic quadrilateral
gives
. Thus, we are to prove that
From the fact that and that
is iscoceles, we find that
. So, we have
. So we are to prove that
We have ,
,
,
,
, and so we are to prove that
We shall show that this is true: Let the altitude from touch
at
. Then it is obvious that
and
and thus
.
Thus we have proven that .
Solution 2
Clearly, is a kite, so its diagonals are perpendicular. Furthermore, we have triangles
and
similar because two corresponding angles are equal.
Hence, we have Notice that we used the fact that a quadrilateral's area is equal to half the product of its perpendicular diagonals (if they are, in fact, perpendicular).
But in (right) triangle , we have
. Furthermore, if
is the intersection of diagonals
and
we have
the midpoint of
and
an altitude of
, so
so
. Hence
as desired.
Solution 3
Proceed as in Solution 2. To prove that , consider that
via usage of definition of sine, equal angles in a right triangle if its altitude is drawn, and the Law of Sines.
See also
1987 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |